Question

If a procedure meets all of the conditions of a binomial distribution except the number of trials is not​ fixed, then the geometric distribution can be used. The probability of getting the first success on the xth trial is given by ​P(x)equalsp (1 minus p )Superscript x minus 1​, where p is the probability of success on any one trial. Subjects are randomly selected for a health survey. The probability that someone is a universal donor​ (with group O and type Rh negative​ blood) is 0.05. Find the probability that the first subject to be a universal blood donor is the fourth person selected.

Answer 1

The probability is 0.0428

Explanation:

First, let's remember that the binomial distribution is given by the formula:

`P(X=k) =\left[\begin{array}{ccc}n\\k\end{array}\right] p^(k)(1-p)^(n-k)` where k is the number of successes in n trials and p is the probability of success.

However, the problem tells us that when there isn't a number of trials fixed, we can use the geometric distribution and the formula for getting the first success on the xth trial becomes:

`P(X=x) = p(1-p)^(x-1)\\`

The problem asks us to find the probability of the first success on the 4th trial (given that the first subject to be a universal blood donor will be the fourth person selected)

Using this formula with the parameters given, we have:

p = 0.05

x = 4

Substituting these parameters in the formula and solving it, we get:

`P(X=4) = 0.05(1-0.05)^(4-1)\\P(X=4) = 0.05 (0.95)^(3)\\P(X=4) = 0.05(.8573)\\P(X=4) = 0.0428`

Therefore, the probability that the first subject to be a universal blood donor is the fourth person selected is 0.0428 or 4.28%